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3 years ago

How many solutions does the equation have?

Mathematics

1 answer:

Lunna [17]3 years ago

6 0

6x + 3 = 3 + 6x
0 = 0

many solutions

answer
infinitely many

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A Freight train is traveling 30 miles per hour. An automobile starts out from the same place 1 hour later and overtakes the trai

Mazyrski [523]

40 I believe because the space of the auto mobile catching up to the train is 4 hours. 4 times 30 is 120 miles so u need to figure out a number that can reach 120 in three hours. 3 times 40 is 120

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2 years ago

Using the diagram above,what is the m

makkiz [27]

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3 years ago

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3 years ago

Read 2 more answers

Does anyone have the questions answered

nata0808 [166]

Answer:

no can you help me

Step-by-step explanation:

6 0

3 years ago

Suppose X has an exponential distribution with mean equal to 23. Determine the following:

e-lub [12.9K]

Answer:

a) P(X > 10) = 0.6473

b) P(X > 20) = 0.4190

c) P(X < 30) = 0.7288

d) x = 68.87

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

Mean equal to 23.

This means that $m = 23, \mu = \frac{1}{23} = 0.0435$

(a) P(X >10)

$P(X > 10) = e^{-0.0435*10} = 0.6473$

P(X > 10) = 0.6473

(b) P(X >20)

$P(X > 20) = e^{-0.0435*20} = 0.4190$

P(X > 20) = 0.4190

(c) P(X <30)

$P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288$

P(X < 30) = 0.7288

(d) Find the value of x such that P(X > x) = 0.05

$P(X > x) = e^{-\mu x}$

$0.05 = e^{-0.0435x}$

$\ln{e^{-0.0435x}} = \ln{0.05}$

$-0.0435x = \ln{0.05}$

$x = -\frac{\ln{0.05}}{0.0435}$

$x = 68.87$

5 0

3 years ago